$\begin {array} {1 1} (A)\;\large\frac{x^2}{25}-\large\frac{y^2}{9}=1 & \quad (B)\;\large\frac{x^2}{9}+\large\frac{y^2}{25}=1 \\ (C)\;\large\frac{y^2}{25}-\large\frac{x^2}{9}=1 & \quad (D)\;\large\frac{x^2}{25}+\large\frac{y^2}{9}=1 \end {array}$

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- Equation of an ellipse whose major axis is along the x - axis is given by $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
- If a point move in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse . The constant value is equal to the length of the major axis of the ellipse.

Step 1 :

Let us take A and B as the two positions of the two flag posts and P be the position of the man.

$ \therefore PA + PB = 10$

Hence it is clear that the path described by the man is an ellipse.

The points A and B are the foci.

The length of the major axis is 10 m.

Step 2 :

The equation of the ellipse is of the form $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$

Since $2a=10 \Rightarrow a = 5$

Distance between the foci is 2c = 8

$ \therefore c = 4$

We know $c = \sqrt{a^2-b^2}$

or $ c^2 = a^2-b^2$

Substituting the value for c and a we get,

$16 = 25-b^2$

$ \therefore b^2=9 \Rightarrow b = 3$

Hence the equation of the path traced by the man is

$\large\frac{x^2}{25}$$+\large\frac{y^2}{9}$$=1$

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